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Name ID w = 3 rad/s 4= 8 rad/s2 At the instant shown, the disk is moving to the left such that it has angular acceleration ão

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\\ \textit{\textbf{Solution:}} \\ \\ \textbf{Part (A) } \\ \textup{Here, the vector radial distance CD is obtained as:} \\ \\ \therefore \vec{r}_{D/C} = r_D-r_C\\ \\ \therefore \vec{r}_{D/C} =( 0.5 \cos 60^o\, i+ 0.5 \sin 60^o\, j) -( 0\, i+ 0\, j)\\ \\ {\color{DarkBlue} \therefore \vec{r}_{D/C} = 0.5 \cos 60^o\, i+ 0.5 \sin 60^o\, j} \\ \\ \\ \textbf{Part (B \& C) } \\ \\ \textup{The vector }\vec{a}_D\times \vec{r}_{D/C}\textup{ and }\vec{\omega}\times (\vec{\omega}\times \vec{r}_{D/C})\textup{ on circle is shown as:}

\\ \textup{ The acceleration of Point C can be obtained as:} \\ \\ \therefore a_C = \alpha r = (0.5) (8) = 4 \, \, m/s^2 =-4\, i\,\, m/s^2 \\ \\ \\ \textbf{Part (D) } \\ \\ \textup{Now, applying the relative motion analysis:}\\ \\ \textup{The acceleration at point D can be written as;}\\ \\ \therefore \vec{a}_D=\vec{a}_C+ \alpha \times \vec{r}_{_D/C}- \vec{\omega}\times (\vec{\omega}\times \vec{r}_{D/C})\\ \\ \therefore \vec{a}_D=\vec{a}_C+ \alpha \times \vec{r}_{_D/C}- \omega^2 \vec{r}_{D/C} \\ \\ \therefore \vec{a}_D = (-4i) + (8k)\times (0.5 \cos 60^o\, i+ 0.5 \sin 60^o\, j) \\ -(3)^2(0.5 \cos 60^o\, i+ 0.5 \sin 60^o\, j) \\ \\ \therefore \vec{a}_D = (-4i) +(2j-3.4641i) -(2.25i+3.89712j)\\ \\ {\color{DarkBlue} \boxed { \therefore \vec{a}_D =(-9.7141\, \, i -1.89712\, \, j ) \, \, m/s^2}} \\

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